1550078481-Ordinary_Differential_Equations__Roberts_

376 Ordinary Differential Equations

A THEOREM ON OBTAINING REAL SOLUTIONS FROM

COMPLEX CONJUGATE EIGENVALUES AND EIGENVECTORS

If A is a constant, real matrix (that is, if A has entries that are all real and

constant) and if,,\= a+i{J where {3 -:f. 0 is an eigenvalue of A with associated

eigenvector x = u + iv, then the linear homogeneous system y' = Ay has

two real-valued, linearly independent solutions of the form

Y1(x) = (e°'x cos{Jx)u - (e°'x sin {Jx)v

Y2(x) = (e°'x sin{Jx)u + (e°'x cos{Jx)v.

Proof: Since,,\= a+i{J is an eigenvalue of A and x = u+iv is an associated

eigenvector, a complex-valued solution of y' = Ay is

(18) z(x) = e(a:+i/J)x(u +iv).

Recall that Euler's formula is eie =cos e + i sine. Therefore,

(19) e(a+i/J)x = e°'x+i/)x = e°'xei/)x = e°'x (cos{Jx + isin{Jx).

Substituting (19) into (18) and multiplying, we find

z(x) = e°'x(cos{Jx + isin{Jx)(u +iv)

= e°'x[(cos{Jx)u-(sin{Jx)v] +ie°'x[(sin{Jx)u+ (cos{Jx)v]

= Y1(x) + iy2(x).

Since z(x) is a complex-valued solution of y' = Ay, we have z'(x) =Azor

(20)

Equating the real parts and the imaginary parts of equation (20), we see

that y~ = Ay 1 and y~ = Ay 2. That is, y 1 and Y2 are both solutions of

y' = Ay. Clearly, t hey are real solutions.

We must still show that the solutions y 1 and y 2 are linearly independent.

By definition, the complex conjugate of ,,\ = a+ i{J is >. = a - i{J. Since

A is assumed to have real, constant entries, the characteristic polynomial

of A will have real coefficients. So by the complex conjugate root the-

orem if ,,\ = a + i{J is a root of the characteristic polynomial of A , then

so is its complex conjugate >. = a - i{J. The following computations show

that the complex conjugate of x , the vector x = u - iv , is an eigenvector of

A associated with the eigenvalue>.. Since,,\ is an eigenvalue of A and xis an